Rozwiązane

kat alfa jest katem ostrym i cos alfa = 2/3 zatem sinus wynosi...



Odpowiedź :

Vivi

[tex]cos\alpha=\frac23\\\\sin^2\alpha+cos^2\alpha=1\\\\sin^2\alpha+(\frac23)^2=1\\\\sin^2\alpha+\frac49=1\\\\sin^2\alpha=1-\frac49\\\\sin^2\alpha=\frac59\\\\sin\alpha=\frac{\sqrt5}3\ \vee\ sin\alpha=-\frac{\sqrt5}3[/tex]

założenie z zadania:

[tex]\alpha\ \in\ (0^o\ ;\ 90^o)[/tex]

stąd:

Odp:

[tex]sin\alpha=\frac{\sqrt5}3[/tex]

Basetla

[tex]cos\alpha = \frac{2}{3}\\sin\alpha = ?\\\\sin^{2}\alpha + cos^{2}\alpha = 1\\\\sin^{2}\alpha = 1 - cos^{2} = 1 - (\frac{2}{3})^{2} = \frac{9}{9}-\frac{4}{9} = \frac{5}{9}\\\\sin\alpha = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{\sqrt{9}}\\\\\boxed{sin\alpha =\frac{\sqrt{5}}{3}}[/tex]

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